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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
dễ mà , chỉ cần đổi trật tự một chút thì bạn có thể nhận ra đây là HĐT bình phương của 1 hiệu
\(A=\left(x+y\right)^2+\left(x-y\right)^2-2x-y\times x+y\)
\(A=\left(x+y\right)^2-2x-y\times x+y+\left(x-y\right)^2\)
\(A=\left(x+y-x-y\right)^2\)
x2(y - z) + y2(z - x) + z2(x - y)
= z2(x - y) + x2 y - x2 z + y2 z - y2 x
= z2(x - y) + (x2 y - y2 x) + (- x2 z + y2 z)
= (x - y)(z2 + xy - zx - zy)
= (x - y)[(z2 - zx) + (xy - zy)]
= (x - y)(z - x)(z -y)
\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)+y^2\left(z-y+y-x\right)\)
\(=x^2\left(y-z\right)-y^2\left(y-z\right)-y^2\left(x-y\right)+z^2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(y-z\right)+\left(z-y\right)\left(y+z\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(y-x\right)\left(x-z\right)\)
\(2\left(x+y\right)\left(x-y\right)-\left(x-y\right)^2+\left(x+y\right)^2-4y^2\)
\(=\left[\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]-4y^2\)
\(=\left(x+y-x+y\right)^2-\left(2y\right)^2\)
\(=\left(2y\right)^2-\left(2y\right)^2=0\)
Sửa:
\(2\left(x+y\right)\left(x-y\right)-\left(x-y\right)^2+\left(x+y\right)^2-4y^2\)
\(=2\left(x^2-y^2\right)-\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)-4y^2\)
\(=2x^2-2y^2-x^2+2xy-y^2+x^2+2xy+y^2-4y^2\)
\(=2x^2-6y^2+4xy\)
\(=2\left(x^2-3y^2+2xy\right)\)
\(=2\left(x^2-3y^2+3xy-xy\right)\)
\(=2\left[x\left(x-y\right)+3y\left(x-y\right)\right]\)
\(=2\left(x-y\right)\left(3y+x\right)\)
x^2-x-y^2-y
=(x^2-y^2)-(x+y)
=(x-y)(x+y)-(x+y)
=(x+y)(x-y-1)
tick nha
x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz
=x^2y+xy^2+xyz+x^2z+xz^2+xyz+y^2z+yz^2
=xy(x+y+z)+zx(x+y+z)+yz(y+z)
=x(y+z)(x+y+z)+yz(y+z)
=(y+z)(x^2+xy+zx+yz)
=(x+y)(y+z)(z+x)
y(x−2z)2+8xyz+x(y−2z)2−2z(x+y)2
=y(x2−4xz+4z2)+8xyz+x(y2−4yz+4z2)−2z(x2+y2+2xy)
=(yx2+xy2)+(4yz2+4xz2)−(2zx2+2zy2+4xyz)
=xy(x+y)+4z2(x+y)−2z(x2+y2+2xy)
=xy(x+y)+4z2(x+y)−2z(x+y)2
=(x+y)(xy+4z2−2xz−2yz)
=(x+y)[y(x−2z)−2z(x−2z)]
=(x+y)(y−2z)(x−2z)
x2 - x - y2 - y
= (x2 - y2) - (x + y)
= (x - y).(x + y) - (x + y)
= (x + y).(x - y - 1)
\(x^2-x-y^2-y\)
\(=x^2-y^2-x-y\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x-y-1\right)\left(x+y\right)\)
`x^2-y^2-2x+2y`
`=(x^2-y^2)-(2x-2y)`
=(x−y)(x+y)−2(x−y)
1+1=2 4567890